Probabilistic Method. Benny Sudakov. Princeton University

Transcription

1 Probabilistic Method Benny Sudakov Princeton University

2 Rough outline The basic Probabilistic method can be described as follows: In order to prove the existence of a combinatorial structure with certain properties, we construct an appropriate probability space and show that a randomly chosen element in this space has desired properties with positive probability.

3 Ramsey theory Of three ordinary adults, two must have the same sex. D.J. Kleitman Ramsey Theory refers to a large body of deep results in mathematics with underlying philosophy: in large systems complete disorder is impossible! Theorem: (Ramsey 1930) k, l there exists N(k, l) such that any two-coloring of the edges of complete graph on N vertices contains either/or Red complete graph of size k Green complete graph of size l

4 Ramsey numbers Definition: R(k, l) is the minimal N so that every red-green edge coloring of K N contains Red complete graph of size k, or Green complete graph of size l Theorem: (Erdős Szekeresh 1935) In particular R(k, l) ( k + l 2 k 1 ) R(k, k) ( 2k 2 k 1 ) 2 2k

5 Proof: part I Induction on k+l. By definition, R(2, l)= l and R(k, 2) = k. Now suppose that Let R(a, b) ( a + b 2 a 1 ), a + b < k + l. N = R(k 1, l) + R(k, l 1) and consider a red-green coloring of the edges of the complete graph K N. Fix some vertex v of K N and let A, B be the set of vertices connected to v by red, green edges respectively. Since A + B = N 1 we have that A R(k 1, l) or B R(k, l 1).

6 Proof: part II If A R(k 1, l), then A must contain either a green clique of size l or a red clique of size k 1 that together with v gives red clique of size k and we are done. The case B R(k, l 1) is similar. By induction hypothesis, this implies R(k, l) N = R(k 1, l) + R(k, l 1) ( ) ( ) k + l 3 k + l 3 + k 2 k 1 ( ) k + l 2 =. k 1

7 Growth rate of R(k, k) Example: k 1 parts of size k 1 Conjecture: (P. Turán) R(k, k) has polynomial growth in k, moreover R(k, k) c k 2

8 Erdős existence argument Theorem: (Erdős 1947) R(k, k) 2 k/2 Proof: Color the edges of the complete graph K N with N = 2 k/2 red and green randomly and independently with probability 1/2. For any set C of k vertices the probability that C spans ) a monochromatic clique is 2 2 ( k ) 2 = 2 1 ( k 2. Since there are ( N k ) possible choices for C, the probability that coloring contains a monochromatic k-clique is at most ( ) N k ) 2 1 ( k 2 N k k! 2k/2+1 2 k2 /2 = 2k/2+1 k! 1

9 Open problem Determine the correct exponent in the bound for R(k, k) Best current estimates k 2 log 2 R(k, k) 2k

10 Large girth and large chromatic number Definitions: The girth g(g) of a graph is the length of the shortest cycle in G. The chromatic number χ(g) is the minimal number of colors which needed to color the vertices of G so that adjacent vertices get different colors. Note: It is easy to color graph with large girth locally using only three colors. Question: If girth of G is large, can it be colored by few colors?

11 Surprising result Theorem: (P. Erdős 1959.) For all k and l there exists a finite graph G with girth at least l and chromatic number at least k. Remark: Explicit constructions of such graphs were not found until only nine years later in 1968 by Lovász.

12 Bound on χ(g) Definition: A set of pairwise nonadjacent vertices of a graph G is called independent. The independence number α(g) is the size of the largest independent set in G. Lemma: For every graph G on n vertices χ(g) n α(g). Proof: Consider the coloring of G into χ(g) colors. Then one of the colors classes has size at least n/χ(g) and its vertices form an independent set. Thus α(g) n/χ(g), as desired.

13 Probabilistic tools Lemma: (Linearity of expectation.) Let X 1, X 2,..., X n be random variables. Then E i X i = i E[X i ]. (No conditions on random variables!) Lemma: (Markov s inequality.) Let X be a non-negative random variable and λ a real number. Then P[X λ] E[X] λ.

14 Proof: part I Fix θ < 1/l. Let n be sufficiently large and G be a random graph G(n, p) with p = 1/n 1 θ. Let X be the number of cycles in G of length at most l. As θ l < 1, by linearity of expectation, E[X] l i=3 By Markov s inequality n i p i ( O n θl) = o(n). P[X n/2] E[X] n/2 = o(1). Set x = 3 p log n, so that P[α(G) x] < ( ) n (1 p) (x 2 ) x ( ne px/2) x = o(1)

15 Proof: part II For large n both of these events have probability less than 1/2. Thus there is a specific graph G with less than n/2 short cycles, i.e., cycles of length at most l, and with α(g) < x 3n 1 θ log n. Remove a vertex from each short cycle of G. This gives G with at least n/2 vertices, girth greater than l and α(g ) α(g). Therefore χ(g ) G α(g ) n/2 3n 1 θ log n = nθ 6 log n k.

16 Set-pair estimate Theorem: (Bollobás 1965.) Let A 1,..., A m and B 1,..., B m be two families of sets such that A i B j = only if i = j. Then m i=1 A i + B i A i 1 1. In particular if A i = a and B i = b, then m ( a+b a ). Example: Let X be a set of size a + b and consider pairs (A i, B i = X A i ) for all A i X of size a. There are ( a+b a the above theorem is tight. ) such pairs, so

17 Proof: part I Let A i = a i, B i = b i and let X = i (A i B i ). Consider a random order π of X and let X i be the event that in this order all the elements of A i precede all those of B i. To compute probability of X i note that there are (a i + b i )! possible orders of element in A i B i and the number of such orders in which all the elements of A i precede all those of B i is exactly a i!b i!. Therefore P[X i ] = a i!b i! (a i + b i )! = a i + b i a i 1.

18 Proof: part II We claim that events X i are pairwise disjoint. Indeed suppose that there is an order of X in which all the elements of A i precede those of B i and all the elements of A j precede all those of B j. W.l.o.g. assume that the last element of A i appear before the last element of A j. Then all the elements of A i precede all those of B j, contradicting the fact that A i B j. Therefore events X i are pairwise disjoint and so we get 1 m i=1 P[X i ] = m i=1 a i + b i a i 1.

19 Sperner s lemma Theorem: (Sperner 1928.) Let A 1,..., A m be a family of subsets of n element set X which is an antichain, i.e., A i A j for all i j. Then Proof: m n n/2. Let B i = X A i and let A i = a i. Then B i = b i = n a i, A i B i is empty but A j B i for all i j. Bollobás theorem 1 m i=1 a i + b i a i 1 = m i=1 n a i Therefore by 1 ( m n n/2 ).

20 Littlewood-Offord problem Theorem: (Erdős 1945.) Let x 1, x 2,..., x n be real numbers such that all x i 1. For every sequence α = (α 1,..., α n ) with α i { 1, +1} let x α = n i=1 α i x i. Then every open interval I in the real line of length 2 contains at most of the numbers x α. ( n n/2 ) Remark: Kleitman (1970) proved that this is still true if x i are vectors in arbitrary normed space.

21 Proof Replacing x i < 0 by x i we can assume that all x i 1. For every α { 1, 1} n let A α be the subset of {1,..., n} containing all 1 i n with α i = 1. Note that if A α A β then α i β i is either 0 or 2. Hence x α x β = i (α i β i )x i = 2 i A β A α x i 2. This implies that {A α x α I} form an antichain and by Sperner s lemma their number is bounded by ( n n/2 ).

22 Explicit constructions Theorem: (Erdős 1947) There is a 2-edge-coloring of complete graph K N, N = 2 k/2 with no monochromatic clique of size k. Problem: (Erdős $100) Find an explicit such coloring. Explicit def = constructible in polynomial time Theorem: (Frankl and Wilson 1981) There is an explicit 2-edge-coloring of log k complete graph K N, N = k c log log k with no monochromatic clique of size k.

23 Bipartite Ramsey Problem: Find large 0, 1 matrix A with no k k homogeneous submatrices. Submatrix def = intersection of k rows and columns Homogeneous def = containing all 0 or all 1 Randomly: There is N N matrix A with k = 2 log 2 N. Explicitly: There is N N matrix A with k = N 1/2. E.g., take [N] = {0, 1} n and define a x,y = x y(mod 2)

24 Breaking 1/2 barrier Theorem: (Barak, Kindler, Shaltiel, S., Wigderson) For every constant δ > 0 there exists a polynomial time computable N N matrix A with 0, 1 entries such that none of its N δ N δ submatrices is homogeneous. Moreover, every N δ N δ submatrix of A has constant proportion of 0 and of 1.